Four Particles of masses 1 kg, 2 kg, 3 kg, and 4 kg are placed at the corners of a square of side 2 m in the x – y plane in anticlockwise sense. If the origin of the co-ordinate system is taken at the mass of 1 kg, the (x, y) co-ordinates of the centre of mass, expressed in 'metre' are
$(3, \frac{7}{5})$

$(2, \frac{7}{5})$

$(1, \frac{7}{5})$

$(4, \frac{7}{5})$

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Solution

The correct option is C $(1, \frac{7}{5})$

The (x, y) co - ordinates of masses 1 kg, 2kg, 3kg and 4kg respectively are
$(x_{1} = 0, y_{1} = 0), (x_{2} = 2 m, y_{2} = 0), (x_{3} = 2 m, y_{3} = 2 m), (x_{4} = 0, y_{4} = 2 m) .$
The (x, y) co-ordinates of the centre of mass are
$x_{C M} = \frac{1 k g \times 0 + 2 k g \times 2 m + 3 k g \times 2 m + 4 k g \times 0}{1 k g + 2 k g + 3 k g + 4 k g} = 1 m$
$y_{C M} = \frac{1 k g \times 0 + 2 k g \times 0 + 3 k g \times 2 m + 4 k g \times 2 m}{1 k g + 2 k g + 3 k g + 4 k g} = \frac{7}{5} m$

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